Chapter 3 Polynomial Ex.3.6 solution.

RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Ex 3.6.

Class 10 Maths Chapter 3 Polynomials Ex 3.5 Solution is provided in this post. Here we have provide the solutions of RBSE Boards Books according to chapter wise.

Chapter 3 Polynomial Ex.3.6 solution.

Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Ex 3.6.

Question 1.

Find the L.C.M. of following expressions
(i) 24x2yz and 27x4y2z2
(ii) x2 – 3x + 2 and x4 + x3 – 6x2
(iii) 2x2 – 8 and x2 – 5x + 6
(iv) x2 – 1, (x2 + 1)(x + 1) and x2 + x – 1
(v) 18(6x4 + x3 – x2) and 45(25x6 + 3x5 – x4)
Solution

Chapter 3 Polynomial Ex.3.6 solution.

Question 2.
Find the H.C.F. of following expressions.
(i) a3b4, ab5, a2b8
(ii) 16x2y2, 48x4z
(iii) x2 – 7x + 12; x2 – 10x + 21 and x2 + 2x – 15
(iv) (x + 3)2(x – 2) and (x + 3)(x – 2)2
(v) 24(6x4 – x3 – 2x2) and 20(6x6 + 5x5 + x4)
Solution

Chapter 3 Polynomial Ex.3.6 solution.

Question 3.
If u(x) = (x – 1)2 and v(x) = (x2 – 1) then, verify the relation L.C.M. × H.C.F. = u(x) × v(x).
Solution
u(x) = (x – 1)2
⇒ u(x) = (x – 1)(x – 1)
and v(x) = (x2 – 1) = (x – 1)(x + 1)
Product of common least powers = (x – 1)
H.C.F. = (x – 1)
Product of highest power of prime factors = (x – 1)2(x + 1)
L.C.M. = (x – 1)2(x + 1)
Test:
u(x) × v(x) = (x – 1)2 × (x2 – 1) = (x – 1)2(x2 – 1)
and H.C.F. × L.C.M. = (x – 1)(x – 1)2(x + 1) = (x – 1)2(x2 – 1)
So, L.C.M. × H.C.F. = w(x) × v(x).
Hence proved.

Chapter 3 Polynomial Ex.3.6 solution.

Question 4.
The product of two expression are (x – 7)(x2 + 8x + 12) if their H.C.F. is (x + 6), then find their L.C.M.
Solution
Given-Product of two expressions = (x – 7)(x2 + 8x + 12)
= (x – 7)[x2 + 6x + 2x + 12]
= (x – 7)[x(x + 6) + 2(x + 6)]
= (x – 7)(x + 2)(x + 6)
and H.C.F. = x + 6
L.C.M. = ?
we know that
H.C.F. × L.C.M. = Product of expressions

Chapter 3 Polynomial Ex.3.6 solution.

Question 5.
H.C. F. and L.C.M of two quadratic expressions are respectively (x – 5) and x3 – 19x – 30, then find both expressions.
Solution
we know that H.C.F. = (x – 5)
and L.C.M. = x3 – 19x – 30
factorize x3 – 19x – 30
then putting x = 1 = (1)3 – 19(1) – 30 ≠ 0
then putting x = -1 = (-1)3 – 19(-1) – 30 ≠ 0
then putting x = 2 = (2)3 – 19(2) – 30 = 8 – 38 – 30 ≠ 0
then putting x = -2 = (-2)3 – 19(-2) – 30 = -8 + 38 – 30 = 0
Putting x = – 2 in expression we get zero.
(x + 2) is a factor of expression

x3 – 19x – 30 = (x + 2)(x2 – 2x – 15) = (x + 2)(x – 5)(x + 3)
Now H.C.F. = (x – 5) and L.C.M. = (x + 2)(x – 5)(x + 3)
⇒ (x – 5) is common factor
⇒ (x – 5) will exists in second expressions
⇒ First expression = (x – 5)(x + 2) = x2 – 3x – 10
and second expression = (x – 5)(x + 3) = x2 – 2x – 15
Hence, x2 – 3x – 10 and x2 – 2x – 15 are required expressions.

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